Is complement of decidable language decidable?
Theorem: A language is decidable iff both it and its complement are recognizable. Moreover, if a language is decidable, then so is its complement, and hence that complement is recognizable. Now suppose a language L and its complement are recognizable. Let A be a recognizer for L, and B for its complement.
Are semi decidable languages closed under complement?
Here we will show that they are closed under union; moreover they are also closed under intersection, however complementation may create a non-semi- decidable language. Theorem 16. The class of semi-decidable languages is closed under union and intersection operations.
What does closed under complement mean?
A class is said to be closed under complement if the complement of any problem in the class is still in the class. Any class which is closed under complement is equal to its complement class.
Is the complement of a decidable language context free?
d) If a language is context free, then its complement is decidable. TRUE; If a language L is context free, then it is decidable. Since the class of decidable languages is closed under complementation, L is decidable.
What is Turing acceptable?
A language which is Turing Recognizable if there is a Machine that will halt and accept only the strings in that language and not in that language, then that TM either rejects, or does not halt at all. A Language is called Turing Decidable if some Turing Machine decides it.
Are regular languages closed under union?
Regular languages are closed under union, concatenation, star, and complementation.
Are recursively enumerable languages closed under complement?
The class of recursively enumerable languages is not closed under complementation, because there are examples of recursively enumerable languages whose complement is not recursively enumerable. Those examples come from languages that are recursively enumerable, but not recursive.
What is the closure of decidable languages?
✦Decidable languages are closed under ∪, °, *, ∩, and complement. ✦Example: Closure under ∪. ✦Need to show that union of 2 decidable L’s is also decidable. Let M1 be a decider for L1 and M2 a decider for L2 A decider M for L1 ∪L2: On input w: 1. Simulate M1 on w. If M1 accepts, then ACCEPT w.
Is the complement of a decidable language recognizable?
Theorem:A language is decidable iff both it and its complement are recognizable. Proof:Surely, a decidable language is recognizable. Moreover, if a language is decidable, then so is its complement, and hence that complement is recognizable.
Are the decidable languages closed under Union and intersection?
Proposition:The decidable languages are closed under union and intersection. Proof:Let L and M be languages that are decided by algorithms A and B respectively. In order to decide their union (or intersection) simply run A and B in parallel on the same given input string until they either accept or reject.
Is recursion closed under complement and set difference?
Recursive (Turing Decidable) languages are closed under following Kleene star, concatenation, union, intersection, complement and set difference. Recursively enumerable language are closed under Kleene star, concatenation, union, intersection. They are NOT closed under complement or set difference.